Chapter 4, Problem 5CT

To determine

To Find:The exact value of five trigonometric function of theangle $\theta$ .

Answer to Problem 5CT

For $0\le \theta \le \frac{\pi }{2}$ ; $\sin \theta =\frac{3}{\sqrt{13}},\cos \theta =\frac{2}{\sqrt{13}},\cot \theta =\frac{2}{3},\sec \theta =\frac{\sqrt{13}}{2},\cos ec\theta =\frac{\sqrt{13}}{3}$ and

For $\pi \le \theta \le \frac{3\pi }{2}$ ; $\sin \theta =-\frac{3}{\sqrt{13}},\cos \theta =-\frac{2}{\sqrt{13}},\cot \theta =\frac{2}{3},\sec \theta =-\frac{\sqrt{13}}{2},\cos ec\theta =-\frac{\sqrt{13}}{3}$ .

Explanation of Solution

Given information:

The given trigonometric function tan $\theta$ = $\frac{3}{2}$ .

Formula used:

The Pythagoras theorem$p^2+b^2=h^2$

Where $p$ is perpendicular, $b$ is base, $h$ is height.

Also by trigonometric formula,$\begin{array}{l}\sin \theta =\frac{p}{h}\\ \cos \theta =\frac{b}{h}\\ \sec \theta =\frac{1}{\cos \theta }\\ \cos ec\theta =\frac{1}{\sin \theta }\\ \cot \theta =\frac{1}{\tan \theta }\end{array}$

Where $p$ is perpendicular, $b$ is base, $h$ is height.

Calculation:

Consider the statement “given trigonometric function $\tan \theta =\frac{3}{2}$ ”

Recall that by Pythagoras formula

  $\begin{array}{l}{p}^2+b^2=h^2\\ 3^2+2^2=h^2\\ 9+4=h^2\\ 13=h^2\\ h=\sqrt{13}\end{array}$

Since from

  $0\le \theta \le \frac{\pi }{2}$ the trigonometric function are positive .

Recall theformula of sinθ=ph .

Where

  $p$ is perpendicular and

  $h$ is height

Therefore

  $\sin \theta =\frac{3}{\sqrt{13}}$ .

Recall theformula of cosθ is $\frac{b}{h}$ .

Where b is base and h is height.

Therefore $\cos \theta =\frac{2}{\sqrt{13}}$ .

Recall the formula of cosecθ is $\frac{1}{\sin \theta }$ .

Therefore $\cos ec\theta =\frac{\sqrt{13}}{3}$ .

Recall the formula of secθ is $\frac{1}{\cos \theta }$ .

Therefore $\sec \theta =\frac{\sqrt{13}}{2}$ .

Recall the formula of cotθ is $\frac{1}{\tan \theta }$ .

Therefore $\cot \theta =\frac{2}{3}$ .

So, the five trigonometric functions are$\begin{array}{l}\sin \theta =\frac{3}{\sqrt{13}}\\ \cos \theta =\frac{2}{\sqrt{13}}\\ \cos ec\theta =\frac{\sqrt{13}}{3}\\ sec\theta =\frac{\sqrt{13}}{2}\\ \cot \theta =\frac{2}{3}\end{array}$

Now from

  $\pi \le \theta \le \frac{3\pi }{2}$ the trigonometric functions are negative except

  $\left(\tan \theta \right)$ and $\left(\cot \theta \right)$ .

So the value of five trigonometric function are:

  $\begin{array}{l}\sin \theta =-\frac{3}{\sqrt{13}}\\ \cos \theta =-\frac{2}{\sqrt{13}}\\ \cot \theta =\frac{2}{3}\\ \sec \theta =-\frac{\sqrt{13}}{2}\\ \cos ec\theta =-\frac{\sqrt{13}}{3}\\ \end{array}$